8 A tractor has a mass of 6000 kg . When developing a power of 5 kW , the tractor is travelling at a steady speed of \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) across a horizontal field.
- Calculate the magnitude of the resistance to the motion of the tractor.
The tractor comes to horizontal ground where the resistance to motion is different. The power developed by the tractor during the next 10 s has an average value of 8 kW . During this time, the tractor accelerates uniformly from \(2.5 \mathrm {~ms} ^ { - 1 }\) to \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
- (A) Show that the work done against the resistance to motion during the 10 s is 71750 J .
(B) Assuming that the resistance to motion is constant, calculate its value.
The tractor can usually travel up a straight track inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac { 1 } { 20 }\), while accelerating uniformly from \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(3.25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) over a distance of 100 m against a resistance to motion of constant magnitude of 2000 N .
The tractor develops a fault which limits its maximum power to 16 kW . - Determine whether the tractor could now perform the same motion up the track.
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[You should assume that the mass of the tractor and the resistance to motion remain the same.]
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\caption{Fig. 9}
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Fig. 9 shows the instant of impact of two identical uniform smooth spheres, A and B , each with mass \(m\). Immediately before they collide, the spheres are sliding towards each other on a smooth horizontal table in the directions shown in the diagram, each with speed \(v\). The coefficient of restitution between the spheres is \(\frac { 1 } { 2 }\). - Show that, immediately after the collision, the speed of A is \(\frac { 1 } { 8 } v\). Find its direction of motion.
- Find the percentage of the original kinetic energy that is lost in the collision.
- State where in your answer to part (i) you have used the assumption that the contact between the spheres is smooth.
In this question take \(g = 10\).
A smooth ball of mass 0.1 kg is projected from a point on smooth horizontal ground with speed \(65 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 3 } { 4 }\). While it is in the air the ball is modelled as a particle moving freely under gravity. The ball bounces on the ground repeatedly. The coefficient of restitution for the first bounce is 0.4 .
\section*{(i) Show that the ball leaves the ground after the first bounce with a horizontal speed of \(52 \mathrm {~ms} ^ { - 1 }\) and a vertical speed of \(15.6 \mathrm {~ms} ^ { - 1 }\). Explain your reasoning carefully.}
\section*{(ii) Calculate the magnitude of the impulse exerted on the ball by the ground at the first bounce.}
Each subsequent bounce is modelled by assuming that the coefficient of restitution is 0.4 and that the bounce takes no time. The ball is in the air for \(T _ { 1 }\) seconds between projection and bouncing the first time, \(T _ { 2 }\) seconds between the first and second bounces, and \(T _ { n }\) seconds between the \(( n - 1 )\) th and \(n\)th bounces.
- (A) Show that \(T _ { 1 } = \frac { 39 } { 5 }\).
(B) Find an expression for \(T _ { n }\) in terms of \(n\). - According to the model, how far does the ball travel horizontally while it is still bouncing?
- According to the model, what is the motion of the ball after it has stopped bouncing?